Difficulty: Easy
Correct Answer: Class C
Explanation:
Introduction / Context:Although modern networks use Classless Inter-Domain Routing (CIDR), legacy classful ranges still appear in exams and historical documentation. Recognizing the host capacity of each class helps map older designs to current subnetting practice.
Given Data / Assumptions:
Concept / Approach:For a /24 network, there are 2^8 = 256 addresses in the host portion. Subtract 2 for network and broadcast, leaving 254 usable hosts. This aligns with the canonical Class C networks (192.0.0.0–223.255.255.255) under classful rules.
Step-by-Step Solution:
Compute host space for Class C: 8 host bits.Total addresses = 2^8 = 256.Usable hosts = 256 − 2 = 254.Verification / Alternative check:Contrast with Class A (/8) and Class B (/16) which provide far more host addresses. Class D is multicast, not for hosts; Class E is experimental.
Why Other Options Are Wrong:
Common Pitfalls:Forgetting to subtract network and broadcast; applying classful logic to CIDR blocks where host counts vary by mask.
Final Answer:Class C
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